# Closed form expression for the gamma function for complex numbers $s=a+ib$ with $b≠0$

Let us consider the famous gamma function.

My question is:

Is it possible to derive a closed expression for the gamma function for complex numbers $s=a+ib$ with $a≠1/2$, $b≠0$.

I know that such a closed form exists for integer values of $x=n$ for which $G(n)=(n-1)!$. Also, some non integers values have a closed form. Also if $a=1/2$ then a closed form exists. See this link:

http://mathoverflow.net/questions/112682/riemann-siegel-function-and-gamma-function

-
You may find expressions for the module for $a=0, a=\frac 12$ or $a=1$ here ($(6.1.29)$ and more) (btw your link shows rather a relation of the argument of zeta with $\zeta(1/2+it)$ : $\Gamma$ is the 'simpler' function!). –  Raymond Manzoni Dec 23 '12 at 9:16
@RaymondManzoni: Thank you very much. –  ZE1 Dec 23 '12 at 9:48
You are welcome ! (sorry for the typos 'module' $\to$ modulus and the MO link gives a relation of the argument of zeta with $\Gamma(1/2+it)$ : zeta is the complicated function and the relation just shows that its argument is simple since it may be written as a function of $\Gamma$ and other elementary functions). $\Gamma$ itself has only trivial closed forms as opposed to the derivative of $\log(\Gamma)$ i.e. the digamma or $\psi$ function that admits a closed form at every rational value. –  Raymond Manzoni Dec 23 '12 at 10:14
I don't know of an impossibility proof. Proofs exist that the gamma function can't be written as an elementary function but this doesn't exclude specific closed forms as at $\frac 12$. –  Raymond Manzoni Dec 23 '12 at 10:32
The closest thing I could find is the Hölder theorem from the discussion here. See too the more recent paper1 in french and paper2. –  Raymond Manzoni Dec 23 '12 at 11:03
show 3 more comments

## 1 Answer

The $\Gamma$ function lacks a closed form containing only elementary functions. Here are the equivalent formulas of the $\Gamma$ function however:

$$\Gamma(z)=\int_{0}^{+\infty}t^{z-1}e^{-t}dt$$ $$\Gamma(z)=\frac1z\prod_{n=1}^{\infty}\frac{(1+\frac1n)^z}{1+\frac zn}$$ $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\frac{e^{\frac zn}}{1+\frac zn}$$ where $\gamma$ is the Euler–Mascheroni constant. Of course we can relate the $\Gamma$ function with elementary functions via the following indentity: $$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$$ We also have Riemann's functional equation $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$ which relates the $\Gamma$ and the $\zeta$ functions.

If by closed form we mean an expression containing only elementary functions then no, $\Gamma$ has no such form. For more information read this

-
I mean this case: $$\Gamma(s)=f(a,b)$$ such that the expression of $f$ is well know. I find one but I want to use it as contraduction in a proof. –  ZE1 Dec 23 '12 at 9:19
The expression of $f$ is given in term of $a$ and $b$ in an explicit manner. –  ZE1 Dec 23 '12 at 9:21
I find this explicit formula: $arg(Γ(α+iβ))=λ(α,β)-σ(α,β)$ with $λ(α,β)$ and $σ(α,β)$ are given in term of $cos,sinh,sin, cosh$. –  ZE1 Dec 23 '12 at 9:29
@user53124 And so? Even if $\arg \Gamma$ is elementary, $\Gamma$ is not elementary so any attempt to find such a form wil be futile. –  Nameless Dec 23 '12 at 9:31
"If by closed form we mean an expression containing only elementary functions then no, $\Gamma$ has no such form" : does not exists or impossible. –  ZE1 Dec 23 '12 at 9:31
show 5 more comments